3.1 \(\int x^3 (a+b x) \sin (c+d x) \, dx\)

Optimal. Leaf size=126 \[ -\frac {6 a \sin (c+d x)}{d^4}+\frac {6 a x \cos (c+d x)}{d^3}+\frac {3 a x^2 \sin (c+d x)}{d^2}-\frac {a x^3 \cos (c+d x)}{d}-\frac {24 b \cos (c+d x)}{d^5}-\frac {24 b x \sin (c+d x)}{d^4}+\frac {12 b x^2 \cos (c+d x)}{d^3}+\frac {4 b x^3 \sin (c+d x)}{d^2}-\frac {b x^4 \cos (c+d x)}{d} \]

[Out]

-24*b*cos(d*x+c)/d^5+6*a*x*cos(d*x+c)/d^3+12*b*x^2*cos(d*x+c)/d^3-a*x^3*cos(d*x+c)/d-b*x^4*cos(d*x+c)/d-6*a*si
n(d*x+c)/d^4-24*b*x*sin(d*x+c)/d^4+3*a*x^2*sin(d*x+c)/d^2+4*b*x^3*sin(d*x+c)/d^2

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Rubi [A]  time = 0.31, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6742, 3296, 2637, 2638} \[ \frac {3 a x^2 \sin (c+d x)}{d^2}-\frac {6 a \sin (c+d x)}{d^4}+\frac {6 a x \cos (c+d x)}{d^3}-\frac {a x^3 \cos (c+d x)}{d}+\frac {4 b x^3 \sin (c+d x)}{d^2}+\frac {12 b x^2 \cos (c+d x)}{d^3}-\frac {24 b x \sin (c+d x)}{d^4}-\frac {24 b \cos (c+d x)}{d^5}-\frac {b x^4 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x)*Sin[c + d*x],x]

[Out]

(-24*b*Cos[c + d*x])/d^5 + (6*a*x*Cos[c + d*x])/d^3 + (12*b*x^2*Cos[c + d*x])/d^3 - (a*x^3*Cos[c + d*x])/d - (
b*x^4*Cos[c + d*x])/d - (6*a*Sin[c + d*x])/d^4 - (24*b*x*Sin[c + d*x])/d^4 + (3*a*x^2*Sin[c + d*x])/d^2 + (4*b
*x^3*Sin[c + d*x])/d^2

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int x^3 (a+b x) \sin (c+d x) \, dx &=\int \left (a x^3 \sin (c+d x)+b x^4 \sin (c+d x)\right ) \, dx\\ &=a \int x^3 \sin (c+d x) \, dx+b \int x^4 \sin (c+d x) \, dx\\ &=-\frac {a x^3 \cos (c+d x)}{d}-\frac {b x^4 \cos (c+d x)}{d}+\frac {(3 a) \int x^2 \cos (c+d x) \, dx}{d}+\frac {(4 b) \int x^3 \cos (c+d x) \, dx}{d}\\ &=-\frac {a x^3 \cos (c+d x)}{d}-\frac {b x^4 \cos (c+d x)}{d}+\frac {3 a x^2 \sin (c+d x)}{d^2}+\frac {4 b x^3 \sin (c+d x)}{d^2}-\frac {(6 a) \int x \sin (c+d x) \, dx}{d^2}-\frac {(12 b) \int x^2 \sin (c+d x) \, dx}{d^2}\\ &=\frac {6 a x \cos (c+d x)}{d^3}+\frac {12 b x^2 \cos (c+d x)}{d^3}-\frac {a x^3 \cos (c+d x)}{d}-\frac {b x^4 \cos (c+d x)}{d}+\frac {3 a x^2 \sin (c+d x)}{d^2}+\frac {4 b x^3 \sin (c+d x)}{d^2}-\frac {(6 a) \int \cos (c+d x) \, dx}{d^3}-\frac {(24 b) \int x \cos (c+d x) \, dx}{d^3}\\ &=\frac {6 a x \cos (c+d x)}{d^3}+\frac {12 b x^2 \cos (c+d x)}{d^3}-\frac {a x^3 \cos (c+d x)}{d}-\frac {b x^4 \cos (c+d x)}{d}-\frac {6 a \sin (c+d x)}{d^4}-\frac {24 b x \sin (c+d x)}{d^4}+\frac {3 a x^2 \sin (c+d x)}{d^2}+\frac {4 b x^3 \sin (c+d x)}{d^2}+\frac {(24 b) \int \sin (c+d x) \, dx}{d^4}\\ &=-\frac {24 b \cos (c+d x)}{d^5}+\frac {6 a x \cos (c+d x)}{d^3}+\frac {12 b x^2 \cos (c+d x)}{d^3}-\frac {a x^3 \cos (c+d x)}{d}-\frac {b x^4 \cos (c+d x)}{d}-\frac {6 a \sin (c+d x)}{d^4}-\frac {24 b x \sin (c+d x)}{d^4}+\frac {3 a x^2 \sin (c+d x)}{d^2}+\frac {4 b x^3 \sin (c+d x)}{d^2}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 82, normalized size = 0.65 \[ \frac {d \left (3 a \left (d^2 x^2-2\right )+4 b x \left (d^2 x^2-6\right )\right ) \sin (c+d x)-\left (a d^2 x \left (d^2 x^2-6\right )+b \left (d^4 x^4-12 d^2 x^2+24\right )\right ) \cos (c+d x)}{d^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x)*Sin[c + d*x],x]

[Out]

(-((a*d^2*x*(-6 + d^2*x^2) + b*(24 - 12*d^2*x^2 + d^4*x^4))*Cos[c + d*x]) + d*(4*b*x*(-6 + d^2*x^2) + 3*a*(-2
+ d^2*x^2))*Sin[c + d*x])/d^5

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fricas [A]  time = 0.55, size = 85, normalized size = 0.67 \[ -\frac {{\left (b d^{4} x^{4} + a d^{4} x^{3} - 12 \, b d^{2} x^{2} - 6 \, a d^{2} x + 24 \, b\right )} \cos \left (d x + c\right ) - {\left (4 \, b d^{3} x^{3} + 3 \, a d^{3} x^{2} - 24 \, b d x - 6 \, a d\right )} \sin \left (d x + c\right )}{d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)*sin(d*x+c),x, algorithm="fricas")

[Out]

-((b*d^4*x^4 + a*d^4*x^3 - 12*b*d^2*x^2 - 6*a*d^2*x + 24*b)*cos(d*x + c) - (4*b*d^3*x^3 + 3*a*d^3*x^2 - 24*b*d
*x - 6*a*d)*sin(d*x + c))/d^5

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giac [A]  time = 0.36, size = 86, normalized size = 0.68 \[ -\frac {{\left (b d^{4} x^{4} + a d^{4} x^{3} - 12 \, b d^{2} x^{2} - 6 \, a d^{2} x + 24 \, b\right )} \cos \left (d x + c\right )}{d^{5}} + \frac {{\left (4 \, b d^{3} x^{3} + 3 \, a d^{3} x^{2} - 24 \, b d x - 6 \, a d\right )} \sin \left (d x + c\right )}{d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)*sin(d*x+c),x, algorithm="giac")

[Out]

-(b*d^4*x^4 + a*d^4*x^3 - 12*b*d^2*x^2 - 6*a*d^2*x + 24*b)*cos(d*x + c)/d^5 + (4*b*d^3*x^3 + 3*a*d^3*x^2 - 24*
b*d*x - 6*a*d)*sin(d*x + c)/d^5

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maple [B]  time = 0.02, size = 359, normalized size = 2.85 \[ \frac {\frac {b \left (-\left (d x +c \right )^{4} \cos \left (d x +c \right )+4 \left (d x +c \right )^{3} \sin \left (d x +c \right )+12 \left (d x +c \right )^{2} \cos \left (d x +c \right )-24 \cos \left (d x +c \right )-24 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}+a \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )-\frac {4 b c \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d}-3 a c \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )+\frac {6 b \,c^{2} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}+3 a \,c^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )-\frac {4 b \,c^{3} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d}+a \,c^{3} \cos \left (d x +c \right )-\frac {b \,c^{4} \cos \left (d x +c \right )}{d}}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x+a)*sin(d*x+c),x)

[Out]

1/d^4*(b/d*(-(d*x+c)^4*cos(d*x+c)+4*(d*x+c)^3*sin(d*x+c)+12*(d*x+c)^2*cos(d*x+c)-24*cos(d*x+c)-24*(d*x+c)*sin(
d*x+c))+a*(-(d*x+c)^3*cos(d*x+c)+3*(d*x+c)^2*sin(d*x+c)-6*sin(d*x+c)+6*(d*x+c)*cos(d*x+c))-4*b*c/d*(-(d*x+c)^3
*cos(d*x+c)+3*(d*x+c)^2*sin(d*x+c)-6*sin(d*x+c)+6*(d*x+c)*cos(d*x+c))-3*a*c*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c
)+2*(d*x+c)*sin(d*x+c))+6/d*b*c^2*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))+3*a*c^2*(sin(d*x+c
)-(d*x+c)*cos(d*x+c))-4/d*b*c^3*(sin(d*x+c)-(d*x+c)*cos(d*x+c))+a*c^3*cos(d*x+c)-1/d*b*c^4*cos(d*x+c))

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maxima [B]  time = 0.54, size = 306, normalized size = 2.43 \[ \frac {a c^{3} \cos \left (d x + c\right ) - \frac {b c^{4} \cos \left (d x + c\right )}{d} - 3 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a c^{2} + \frac {4 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c^{3}}{d} + 3 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} a c - \frac {6 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b c^{2}}{d} - {\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} a + \frac {4 \, {\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b c}{d} - \frac {{\left ({\left ({\left (d x + c\right )}^{4} - 12 \, {\left (d x + c\right )}^{2} + 24\right )} \cos \left (d x + c\right ) - 4 \, {\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \sin \left (d x + c\right )\right )} b}{d}}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)*sin(d*x+c),x, algorithm="maxima")

[Out]

(a*c^3*cos(d*x + c) - b*c^4*cos(d*x + c)/d - 3*((d*x + c)*cos(d*x + c) - sin(d*x + c))*a*c^2 + 4*((d*x + c)*co
s(d*x + c) - sin(d*x + c))*b*c^3/d + 3*(((d*x + c)^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*a*c - 6*(((
d*x + c)^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*b*c^2/d - (((d*x + c)^3 - 6*d*x - 6*c)*cos(d*x + c) -
3*((d*x + c)^2 - 2)*sin(d*x + c))*a + 4*(((d*x + c)^3 - 6*d*x - 6*c)*cos(d*x + c) - 3*((d*x + c)^2 - 2)*sin(d
*x + c))*b*c/d - (((d*x + c)^4 - 12*(d*x + c)^2 + 24)*cos(d*x + c) - 4*((d*x + c)^3 - 6*d*x - 6*c)*sin(d*x + c
))*b/d)/d^4

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mupad [B]  time = 0.28, size = 122, normalized size = 0.97 \[ \frac {6\,a\,x\,\cos \left (c+d\,x\right )+12\,b\,x^2\,\cos \left (c+d\,x\right )}{d^3}-\frac {6\,a\,\sin \left (c+d\,x\right )+24\,b\,x\,\sin \left (c+d\,x\right )}{d^4}-\frac {a\,x^3\,\cos \left (c+d\,x\right )+b\,x^4\,\cos \left (c+d\,x\right )}{d}+\frac {3\,a\,x^2\,\sin \left (c+d\,x\right )+4\,b\,x^3\,\sin \left (c+d\,x\right )}{d^2}-\frac {24\,b\,\cos \left (c+d\,x\right )}{d^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sin(c + d*x)*(a + b*x),x)

[Out]

(6*a*x*cos(c + d*x) + 12*b*x^2*cos(c + d*x))/d^3 - (6*a*sin(c + d*x) + 24*b*x*sin(c + d*x))/d^4 - (a*x^3*cos(c
+ d*x) + b*x^4*cos(c + d*x))/d + (3*a*x^2*sin(c + d*x) + 4*b*x^3*sin(c + d*x))/d^2 - (24*b*cos(c + d*x))/d^5

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sympy [A]  time = 2.34, size = 151, normalized size = 1.20 \[ \begin {cases} - \frac {a x^{3} \cos {\left (c + d x \right )}}{d} + \frac {3 a x^{2} \sin {\left (c + d x \right )}}{d^{2}} + \frac {6 a x \cos {\left (c + d x \right )}}{d^{3}} - \frac {6 a \sin {\left (c + d x \right )}}{d^{4}} - \frac {b x^{4} \cos {\left (c + d x \right )}}{d} + \frac {4 b x^{3} \sin {\left (c + d x \right )}}{d^{2}} + \frac {12 b x^{2} \cos {\left (c + d x \right )}}{d^{3}} - \frac {24 b x \sin {\left (c + d x \right )}}{d^{4}} - \frac {24 b \cos {\left (c + d x \right )}}{d^{5}} & \text {for}\: d \neq 0 \\\left (\frac {a x^{4}}{4} + \frac {b x^{5}}{5}\right ) \sin {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x+a)*sin(d*x+c),x)

[Out]

Piecewise((-a*x**3*cos(c + d*x)/d + 3*a*x**2*sin(c + d*x)/d**2 + 6*a*x*cos(c + d*x)/d**3 - 6*a*sin(c + d*x)/d*
*4 - b*x**4*cos(c + d*x)/d + 4*b*x**3*sin(c + d*x)/d**2 + 12*b*x**2*cos(c + d*x)/d**3 - 24*b*x*sin(c + d*x)/d*
*4 - 24*b*cos(c + d*x)/d**5, Ne(d, 0)), ((a*x**4/4 + b*x**5/5)*sin(c), True))

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